Question

the volume of a cone of radius r and height h is given by v=1/3pir^2h. If the radius and height is both increasing at a constant rate of 1/2 centimeter per second, at what rate, in cubic centimeters per second, is the volume increasing when the height is 9 centimeters and the radius is 6 centimeters

Answer 1

Answer:

The rate of change of the volume $\frac{dV}{dt}$ when the height is 9 centimeters and the radius is 6 centimeters is $24\pi \:\frac{cm^3}{s}$

Step-by-step explanation:

This is a related rate problem because you know a rate and want to find another rate that is related to it. If 2 variables both vary with respect to time and have a relation between them, we can express the rate of change of one in terms of the other.

From the information given we know:

• $\frac{dr}{dt}=\frac{1}{2}\:\frac{cm}{s}$
• $\frac{dh}{dt}=\frac{1}{2}\:\frac{cm}{s}$
• The volume of a cone of radius r and height h is given by $V=\frac{1}{3} \pi r^2 h$

We want to find the rate of change of the volume $\frac{dV}{dt}$ when the height is 9 centimeters and the radius is 6 centimeters.

Applying implicit differentiation to the formula of the volume of a cone we get

$\frac{dV}{dt}=\frac{1}{3}\pi [r^2\frac{dh}{dt}+2rh\frac{dr}{dt} ]$

Substituting the values we know into the above formula:

$\frac{dV}{dt}=\frac{1}{3}\pi [(6)^2\frac{1}{2}+2(6)(9)\frac{1}{2} ]\\\\\frac{dV}{dt}=\frac{1}{3}\pi[18+54]\\\\\frac{dV}{dt}=\frac{72\pi}{3}=24\pi \:\frac{cm^3}{s}$